Q14 / 48 · JEE Advanced 2025 (Paper 2) · Chemistry
An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K . Its cell potential is $\frac{\boldsymbol{X}}{\boldsymbol{F}} \times 10^3$ volts, where $\boldsymbol{F}$ is the Faraday constant. The value of $\boldsymbol{X}$ is $\qquad$ .
Use : Standard Gibbs energies of formation at 298 K are : $\Delta_f G_{\mathrm{CO}_2}^{\circ}=-394 \mathrm{~kJ} \mathrm{~mol}^{-1}$;
$\Delta_f G_{\text {water }}^{\circ}=-237 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \Delta_f G_{\text {butane }}^{\circ}=-18 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Use : Standard Gibbs energies of formation at 298 K are : $\Delta_f G_{\mathrm{CO}_2}^{\circ}=-394 \mathrm{~kJ} \mathrm{~mol}^{-1}$; $\Delta_f G_{\text {water }}^{\circ}=-237 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \Delta_f G_{\text {butane }}^{\circ}=-18 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Correct answer: 105.5
Solution
$\begin{aligned} & \mathrm{C}_4 \mathrm{H}_{10}(\mathrm{~g})+\frac{13}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_2(\mathrm{~g})+5 \mathrm{H}_2 \mathrm{O}(l) \\ & \Delta_{\mathrm{r}} \mathrm{G}^{\circ}=4 \Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{CO}_2}^{\circ}+5 \Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{H}_2 \mathrm{O}}^{\circ}-\Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{C}_4 \mathrm{H}_{10}}^{\circ} \\ & =4 \times(-394)+5(-237)+18 \\ & =-2743 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta_{\mathrm{r}} \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ} \\ & -2743 \times 1000=-26 \times \mathrm{FE}^{\circ} \\ & \mathrm{E}^{\circ}=\frac{105.5}{\mathrm{~F}} \times 10^3=105.50\end{aligned}$
Hence, the correct answer is 105.5.